Page 93 - Calc for the Clueless

Example 4—

2
Find the point c in the mean value theorem or explain why it does not exist if f(x) = x + 3x + 5 on [1,4].
Polynomials are always continuous and differentiable everywhere. The MVT holds, f'(x) = 2x + 3. f'(c) = 2c +
3. Therefore

Notice that 2.5 is between 1 and 4. Let's try others.

Example 5—

Same as above if

The MVT does not hold since the function is not continuous at x = 3.

Example 6—

Same as above except f(x) = (x - 4) , [0,10].
1/3

2/3
f'(x) = 1/[3(x-4) ] and the derivative does not exist at x = 4. The mean value theorem again does not hold.
Example 7—

Same as above except f(x) = x [0,8].
1/3
f'(x) = 1/(3x ). The function is continuous. The derivative exists at every point except x = 0. The MVT holds
2/3
because f'(x) does not have to exist at the end points.

Note

You alert students might have seen in Example 6 that the MVT does not hold, yet there is a point that satisfies
the theorem. This is possible. If the MVT holds, such a point c is guaranteed. If the MVT doesn't hold, such a
point is possible but not guaranteed.

Approximations, Approximations

In an age where computers and calculators, even those fun graphing calculators, do so many things, some things
cannot be done exactly.

We know that all quadratics can be solved using the quadratic formula. Similarly, there is a cubic formula and a
quartic (fourth-degree) formula that can solve all cubics or quartics (although they are truly ugly and messy).
However, in higher mathematics we can prove that most general fifth-degree equations cannot be solved. More
simply, an equation like 2x = cos x cannot be solved exactly. However, we can approximate a solution very
closely.